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\begin{remark*}
    ``Exponentials beat polynomials''
    \[ \lim_{x \to \infty} \frac{e^x}{x^k} =
    \infty \]
    ($k > 0$). This is easy to prove since
    \[ e^x = \sum_{j = 0}^\infty \frac{x^j}{j!} >
    \frac{x^n}{n!} \]
    for $x > 0$. Now pick $n > k$ and then
    \[ \frac{e^x}{x^k} > \frac{x^{n - k}}{n!} \to
    \infty \]
    as $x \to \infty$.
\end{remark*}

\subsubsection*{Trigonometric Functions}

\[ \cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} -
\cdots = \sum_{n = 0}^\infty \frac{(-1)^k
z^{2k}}{(2k)!} \]
\[ \sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} -
\cdots = \sum_{n = 0}^\infty \frac{(-1)^k z^{2k +
1}}{(2k + 1)!} \]
Both power series have infinite radius of
convergence and by Theorem 4.4 we get
\[ (\sin z)' = \cos z, \qquad (\cos z)' = -\sin z \]
($e^z = e(z)$)
\[ e^{iz} = \sum_{n = 0}^\infty \frac{(iz)^n}{n!}
= \sum_{n = 0}^\infty \frac{(iz)^{2k}}{(2k)!} +
\sum_{n = 0}^\infty \frac{(iz)^{2k + 1}}{(2k +
1)!} \]
\[ (iz)^{2k} = (-1)^k z^{2k}, \qquad (iz)^{2k + 1}
= i(-1)^{k} z^{2k + 1} \]
\[ \implies e^{iz} = \cos z + i \sin z \]
Similarly
\[ e^{-iz} = \cos z - i \sin z \]
which gives
\[ \cos z = \half (e^{iz} + e^{-iz}) \]
\[ \sin z = \frac{1}{2i} (e^{iz} - e^{-iz}) \]
From this we get many trigonometric identities:
\[ \cos z = \cos(-z), \qquad \sin (-z) = -\sin z \]
\[ \cos(0) = 1, \qquad \sin (0) = 0 \]
Addition formulas:
\begin{enumerate}[(1)]
    \item $\sin(z + w) = \sin z \cos w + \cos z +
        \sin w$
    \item $\cos(z + w) = \cos z \cos w - \sin z
        \sin w$, $z, w \in \CC$.
\end{enumerate}
These follow from $e^{a + b} = e^a e^b$. To prove
(2) write
\begin{align*}
    \cos(z + w)
    &= \half (e^{i(z + w)} + e^{-i(z + w)}) \\
    &= \half (e^{iz} e^{iw} + e^{iz} e^{iw}) \\
    \cos z \cos w - \sin z \sin w
    &= \frac{1}{4} (e^{iz} + e^{-iz})(e^{iw} +
    e^{-iw}) + \frac{1}{4} (e^{iz} -
    e^{-iz})(e^{iw} - e^{-iw})
\end{align*}
operate to get the result. Also we can easily
deduce that $\sin^2 z + \cos^2 z = 1$ for all $z
\in \CC$. Now if $x \in \RR$, then $\sin x, \cos x
\in \RR$ and so $|\sin x|, |\cos x| \le 1$ for $x
\in \RR$.

\begin{remark*}
    They are not bounded over $\CC$. For example
    take
    \[ \cos(iy) = \half (e^{-y} + e^y) \]
    ($y \in \RR$) then as $y \to \infty$,
    $\cos(iy) \to \infty$!
\end{remark*}

\subsubsection*{Periodicity of the Trigonmetric Functions}

\begin{proposition}
    There is a smallest positive number $\omega$
    (where $\sqrt{2} < \frac{\omega}{2} <
    \sqrt{3}$) such that
    \[ \cos \frac{\omega}{2} = 0 \]
\end{proposition}

\begin{proof}
    If $0 < x < 2$ then
    \[ \sin x = \ub{\left( x - \frac{x^3}{3!}
    \right)}_{>0} + \ub{\left( \frac{x^5}{5!} -
    \frac{x^7}{7!} \right)}_{>0} + \cdots \]
    (If $0 < x < 2$ then $\frac{x^{2n - 1}}{(2n -
    1)!} > \frac{x^{2n + 1}}{(2n + 1)!}$) Hence
    $\sin x > 0$. Since $(\cos x)' = -\sin x < 0$
    for $0 < x < 2$, $\cos x$ is strictly
    decreasing. We'll show that $\cos \sqrt{2} >
    0$ and $\cos \sqrt{3} < 0$. Then by the
    intermediate value theorem the existence of
    $\omega$ follows.
    \[ \cos \sqrt{2} = \ub{\left(
    \frac{(\sqrt{2})^4}{4!} -
    \frac{(\sqrt{2})^6}{6!} \right)}_{>0} +
    \ub{(\cdots)}_{>0} + \ub{(\cdots)}_{>0} + \cdots \]
    So $\cos \sqrt{2} > 0$. Now note that
    \[ \cos\sqrt{3} = 1 - \frac{x^2}{2!} +
    \frac{x^4}{4!} - \ub{\left( \frac{x^6}{6!} -
    \frac{x^8}{8!} \right)}_{>0} - \cdots \]
    But
    \[ 1 - \frac{3}{2} + \frac{9}{4 \times 3
    \times 2} = 1 - \frac{3}{2} + \frac{3}{8} =
    -\frac{1}{8} < 0 \]
    so $\cos \sqrt{3} < 0$.
\end{proof}

\begin{corollary}
    $\sin \frac{\omega}{2} = 1$.
\end{corollary}

\begin{proof}
    Use $\sin^2 \frac{\omega}{2} + \cos^2
    \frac{\omega}{2} = 1$ and $\sin
    \frac{\omega}{2} > 0$.
\end{proof}

\myskip
Now define $\pi = \omega$.

\begin{theorem}
    \begin{enumerate}[(1)]
        \item $\sin \left( z + \frac{\pi}{2}
            \right) = \cos z$, $\cos \left( z +
            \frac{\pi}{2} \right) = -\sin z$.
        \item $\sin(z + \pi) = -\sin z$, $\cos(z +
            \pi) = -\cos z$.
        \item $\sin(z + 2\pi) = \sin z$, $\cos(z +
            2\pi) = \cos z$.
    \end{enumerate}
\end{theorem}

\begin{proof}
    Immediate from addition formulas and $\cos
    \frac{\pi}{2} = 0$, $\sin \frac{\pi}{2} = 1$.
\end{proof}

\myskip
This implies
\begin{align*}
    e^{iz + 2\pi i}
    &= \cos(z + 2\pi) + i\sin(z + 2\pi) \\
    &= \cos z + i \sin z \\
    e^{iz}
\end{align*}
so $e^z$ is periodic with period $2\pi i$.

\begin{remark*}
    ``Relation with geometry'' \\
    Given two vectors $x, y \in \RR^2$ define $x
    \cdot y$ as in vectors and matrices:
    \[ x \cdot y = x_1y_2 + x_2y_2 \]
    \[ x = (x_1, x_2) \qquad y = (y_1, y_2) \]
    Cauchy-Schwarz:
    \[ |x \cdot y| \le \lVert x \rVert \lVert y
    \rVert \]
    where $\lVert x \rVert^2 = x_1^2 + x_2^2$.
    So, for $x \neq 0$, $y \neq 0$
    \[ -1 \le \frac{x \cdot y}{\lVert x \rVert
    \lVert y \rVert} \le 1 \]
    Define the angle between $x$ and $y$ as the
    unique $\theta \in [0, \pi]$ such that
    \[ \cos \theta = \frac{x \cdot y}{\lVert x
    \rVert \lVert y \rVert} \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/9874854ea21111ec.png}
    \end{center}
    $x = (h, v)$, $\cos \theta = x \cdot e_1 = h$
\end{remark*}
